Convert to a integer

mark123_

I would like to convert a string to a integer with leading zeros and keep the leading zeros is that possible to do in php. And if so how would I go about that.

laserlight

Don't convert the string in the first place, until you need it 🙂

bradgrafelman

In other words, PHP is a loosely-typed language, so a string is often just as good as an integer and vice versa. Example:

$a = '0032';
$b = $a + 32;
echo "Result is: $b"; // Result is: 64

If you don't like that answer, then consider this: 32 is the same as 032 which is the same as 0032 which is the same as... etc. The difference between '32' and '032' would only be observed in the displaying of that number (e.g. outputting a string). To that end, if you want to display a number left-padded with zeros, consider using [man]printf/man (or one of its numerous relatives). Example:

$a = '0032';
$b = $a + 32;
printf("Result is: %04d", $b); // Result is: 0064

laserlight

bradgrafelman wrote:
If you don't like that answer, then consider this: 32 is the same as 032 which is the same as 0032 which is the same as... etc.

A catch though: 32 is not the same as 032 which is the same as 0032, but that's because the first 0 for the latter two is a prefix to designate them as being in octal representation.

bradgrafelman

laserlight;10971715 wrote:
A catch though: 32 is not the same as 032 which is the same as 0032, but that's because the first 0 for the latter two is a prefix to designate them as being in octal representation.

Eh... true, that's how PHP (and/or most languages/standards) interpret it - guess I should have been more explicit in what I meant.

What I meant to portray is that the decimal number '32' when stored by a computer (read: binary data) is 0b100000 (take that, octal notation! :p). Left-padding zeros doesn't change the value - it's still 32 (mostly because you're not actually adding the zeros - they're already there in the memory/register location).

mark123_

bradgrafelman;10971714 wrote:
In other words, PHP is a loosely-typed language, so a string is often just as good as an integer and vice versa. Example:
$a = '0032';
$b = $a + 32;
echo "Result is: $b"; // Result is: 64
If you don't like that answer, then consider this: 32 is the same as 032 which is the same as 0032 which is the same as... etc. The difference between '32' and '032' would only be observed in the displaying of that number (e.g. outputting a string). To that end, if you want to display a number left-padded with zeros, consider using [man]printf/man (or one of its numerous relatives). Example:
$a = '0032';
$b = $a + 32;
printf("Result is: %04d", $b); // Result is: 0064

Thank you I never thought about printf() I was using settype that didnt seem to work and then i tried $a =(int)$x; and that did not work either. So once again thank you.

mark123_

I tried that but now I am receiving a parse error

foreach ($line as $line => $values){
if (preg_match($string,$values)!=false){
	$num = preg_replace('/[^0-9]/','',$values);
	$n = wordwrap($num, 3,"<br>", true);
	$n = explode(':',$n);
	foreach ($n as $int) {	
	$int = (int)$int;		

printf $int."<br>";
 }

laserlight

Unlike print, printf is a function, so that should be:

printf('&#37;s<br>', $int);

But that would be effectively the same as writing:

echo $int . '<br>';

You would need to decide how exactly you want to format it. You will not be keeping the original number of leading 0s, except by coincidence.

mark123_

When i do it that way i actually lose my leading zeros and on the conversion $int = (int)$int; I lose my values as well.

bradgrafelman

Read the manual for [man]sprintf/man; the "%04d" in my example above is a "conversion specification" - in other words, it's a placeholder that tells PHP you want to output that variable.

EDIT: Alternatively, if you simply want to output it the way it appears in the original data, remove the cast to (int) as laserlight already suggested in post #2 above.